Find the first circular tour that visits all petrol pumps

Suppose there is a circle. There are n petrol pumps on that circle. You are given two sets of data.

1. The amount of petrol that every petrol pump has.2. Distance from that petrol pump to the next petrol pump.

Calculate the first point from where a truck will be able to complete the circle (The truck will stop at each petrol pump and it has infinite capacity). Expected time complexity is O(n). Assume for 1 litre petrol, the truck can go 1 unit of distance.

For example, let there be 4 petrol pumps with amount of petrol and distance to next petrol pump value pairs as {4, 6}, {6, 5}, {7, 3} and {4, 5}. The first point from where truck can make a circular tour is 2nd petrol pump. Output should be “start = 1” (index of 2nd petrol pump).

A Simple Solution is to consider every petrol pumps as starting point and see if there is a possible tour. If we find a starting point with feasible solution, we return that starting point. The worst case time complexity of this solution is O(n^2).

We can use a Queue to store the current tour. We first enqueue first petrol pump to the queue, we keep enqueueing petrol pumps till we either complete the tour, or current amount of petrol becomes negative. If the amount becomes negative, then we keep dequeueing petrol pumps till the current amount becomes positive or queue becomes empty.

Instead of creating a separate queue, we use the given array itself as queue. We maintain two index variables start and end that represent rear and front of queue.

C/C++

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// C program to find circular tour for a truck

#include <stdio.h>

// A petrol pump has petrol and distance to next petrol pump

structpetrolPump

{

intpetrol;

intdistance;

};

// The function returns starting point if there is a possible solution,

// otherwise returns -1

intprintTour(structpetrolPump arr[], intn)

{

// Consider first petrol pump as a starting point

intstart = 0;

intend = 1;

intcurr_petrol = arr[start].petrol - arr[start].distance;

/* Run a loop while all petrol pumps are not visited.

And we have reached first petrol pump again with 0 or more petrol */

while(end != start || curr_petrol < 0)

{

// If curremt amount of petrol in truck becomes less than 0, then

// remove the starting petrol pump from tour

while(curr_petrol < 0 && start != end)

{

// Remove starting petrol pump. Change start

curr_petrol -= arr[start].petrol - arr[start].distance;

start = (start + 1)%n;

// If 0 is being considered as start again, then there is no

// possible solution

if(start == 0)

return-1;

}

// Add a petrol pump to current tour

curr_petrol += arr[end].petrol - arr[end].distance;

end = (end + 1)%n;

}

// Return starting point

returnstart;

}

// Driver program to test above functions

intmain()

{

structpetrolPump arr[] = {{6, 4}, {3, 6}, {7, 3}};

intn = sizeof(arr)/sizeof(arr[0]);

intstart = printTour(arr, n);

(start == -1)? printf("No solution"): printf("Start = %d", start);

return0;

}

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Java

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//Java program to find circular tour for a truck

publicclassPetrol

{

// A petrol pump has petrol and distance to next petrol pump

staticclasspetrolPump

{

intpetrol;

intdistance;

// constructor

publicpetrolPump(intpetrol, intdistance)

{

this.petrol = petrol;

this.distance = distance;

}

}

// The function returns starting point if there is a possible solution,

// otherwise returns -1

staticintprintTour(petrolPump arr[], intn)

{

intstart = 0;

intend = 1;

intcurr_petrol = arr[start].petrol - arr[start].distance;

// If current amount of petrol in truck becomes less than 0, then

// remove the starting petrol pump from tour

while(end != start || curr_petrol < 0)

{

// If current amount of petrol in truck becomes less than 0, then

// remove the starting petrol pump from tour

while(curr_petrol < 0&& start != end)

{

// Remove starting petrol pump. Change start

curr_petrol -= arr[start].petrol - arr[start].distance;

start = (start + 1) % n;

// If 0 is being considered as start again, then there is no

// possible solution

if(start == 0)

return-1;

}

// Add a petrol pump to current tour

curr_petrol += arr[end].petrol - arr[end].distance;

end = (end + 1)%n;

}

// Return starting point

returnstart;

}

// Driver program to test above functions

publicstaticvoidmain(String[] args)

{

petrolPump[] arr = {newpetrolPump(6, 4),

newpetrolPump(3, 6),

newpetrolPump(7, 3)};

intstart = printTour(arr, arr.length);

System.out.println(start == -1? "No Solution": "Start = "+ start);

}

}

//This code is contributed by Sumit Ghosh

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Python

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# Python program to find circular tour for a track

# A petrol pump has petrol and distance to next petrol pimp

classPetrolPump:

# Constructor to create a new node

def__init__(self,petrol, distance):

self.petrol =petrol

self.distance =distance

# The funtion return starting point if there is a possible

# solution otherwise returns -1

defprintTour(arr):

n =len(arr)

# Consider first petrol pump as starting point

start =0

end =1

curr_petrol =arr[start].petrol -arr[start].distance

# Run a loop whie all petrol pumps are not visited

# And we have reached first petrol pump again with 0

# or more petrol

while(end !=start orcurr_petrol < 0):

# If current amount of petrol pumps are not visited

# And we have reached first petrol pump again with

# 0 or more petrol

while(curr_petrol < 0andstart !=end):

# Remove starting petrol pump. Change start

curr_petrol -=arr[start].petrol -arr[start].distance

start =(start +1)%n

# If 0 is being considered as start again, then

# there is no possible solution

ifstart ==0:

return-1

# Add a petrol pump to current tour

curr_petrol +=arr[end].petrol -arr[end].distance

end =(end +1) %n

returnstart

# Driver program to test above function

arr =[PetrolPump(6,4), PetrolPump(3,6), PetrolPump(7,3)]

start =printTour(arr)

print"No solution"ifstart ==-1else"start =", start

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

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C#

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// C# program to find circular

// tour for a truck

usingSystem;

classGFG

{

// A petrol pump has petrol and

// distance to next petrol pump

publicclasspetrolPump

{

publicintpetrol;

publicintdistance;

// constructor

publicpetrolPump(intpetrol,

intdistance)

{

this.petrol = petrol;

this.distance = distance;

}

}

// The function returns starting point

// if there is a possible solution,

// otherwise returns -1

publicstaticintprintTour(petrolPump[] arr,

intn)

{

intstart = 0;

intend = 1;

intcurr_petrol = arr[start].petrol -

arr[start].distance;

// If current amount of petrol in

// truck becomes less than 0, then

// remove the starting petrol pump from tour

while(end != start || curr_petrol < 0)

{

// If current amount of petrol in

// truck becomes less than 0, then

// remove the starting petrol pump from tour

while(curr_petrol < 0 && start != end)

{

// Remove starting petrol pump.

// Change start

curr_petrol -= arr[start].petrol -

arr[start].distance;

start = (start + 1) % n;

// If 0 is being considered as

// start again, then there is no

// possible solution

if(start == 0)

{

return-1;

}

}

// Add a petrol pump to current tour

curr_petrol += arr[end].petrol -

arr[end].distance;

end = (end + 1) % n;

}

// Return starting point

returnstart;

}

// Driver Code

publicstaticvoidMain(string[] args)

{

petrolPump[] arr = newpetrolPump[]

{

newpetrolPump(6, 4),

newpetrolPump(3, 6),

newpetrolPump(7, 3)

};

intstart = printTour(arr, arr.Length);

Console.WriteLine(start == -1 ? "No Solution":

"Start = "+ start);

}

}

// This code is contributed by Shrikant13

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Output:

start = 2

Time Complexity: Seems to be more than linear at first look. If we consider the items between start and end as part of a circular queue, we can observe that every item is enqueued at most two times to the queue. The total number of operations is proportional to total number of enqueue operations. Therefore the time complexity is O(n).